Calculate the value of Kc for the reaction.

Question

Ammonia can be oxidized in oxygen as shown: 

4 NH₃(g) + 3 O₂(g) <=> 2 N₂(g) + 6 H₂O(g). 

When 0.695 mol of NH₃ and 0.551 mol of O₂ are placed in a 1.00 L container at a certain temperature, the equilibrium {N₂} is 0.1165. Calculate the value of K_c for the reaction. 

the answer is 0.0102

please explain

Answer 1

4NH₃(g) + 3O₂(g) <==> 2N₂(g) + 6H₂O(g)

0.695......0.551..............0..............0........Initial

-4x.......-3x.................+2x.........+6x.........Change

0.695-4x...0.5551-3x....0.1165..0.3495...Equilibrium

Since you are told that [N2] at equilibrium = 0.1165, this must be equal to 2x and thus x = 0.05825.

The [H₂O] comes from it being 3x the [N2] according to the balanced equation

[NH₃] = 0.695 - 4x0.05825 = 0.462

[O₂] = 0.551 - 3x0.05825 = 0.37625

[N₂] = 0.1165

[H₂O] = 0.3495

K = [H₂O]⁶[N₂]² / [NH₃]⁴[O₂]³

K = (0.3495)⁶(0.1165)² / (0.462)⁴(0.376)³ = (0.001823)(0.0136) / (0.0456)(0.0532)

K = 0.0102 <--answer